Sequences for Pythagorean Multiples

It has been shown (Between_two_Rational_Numbers on Wiki) that between two real numbers (ℝ) there exists an irrational number. Accordingly, it will be shown here that between two adjacent natural numbers (ℕ) there exists an irrational number (ℝ\ℚ). In addition, an arithmetic progression or sequence of irrational numbers will be generated where the initial common difference (d) between two irrational numbers generated from √xy is an irrational number between √2 < d < √6 which approaches d=1 as the number of natural numbers grows without bound. Therefore, we can proceed as follows:

If x and y are consecutive natural numbers where y = x + 1 then we can show by induction that:

x < √xy < y

where x,y ∈ ℕ and √xy is an irrational number.

First substituting x + 1 for y

x < √x(x + 1) < x+1 (1)

then substituting x = 1 in the basis step

1 < √1(2) < 2 which is true.

Squaring the left, center and right arts of the inequality (1)

x² < (x(x + 1) < (x+1)² (2)

To prove the chain notation for all numbers we add the equation (2x + 1) to all three parts in the inequality as follows:

x² + 2x + 1 < x(x + 1) + 2x + 1 < (x+1)² + 2x + 1 (3)

and then expanding

x² + 2x + 1 < x² + x + 2x + 1 < x² + 2x + 1 + 2x + 1 (4)

adding up all the terms

x² + 2x + 1 < x² + 3x + 1 < x² + 4x + 2 (5)

since

x² + 3x + 1 < x² + 3x + 2 and

x² + 4x + 2 < x² + 4x + 4 then

x² + 2x + 1 < x² + 3x + 1 < x² + 3x + 2 < x² + 4x + 2 < x² + 4x + 4 (6)

which can be reduced to

x² + 2x + 1 < x² + 3x + 2 < x² + 4x + 4 (7)

Factoring

(x + 1)² < (x + 1)(x + 2) < (x + 2)² (8)

where it can be seen that left part < middle part < right part. Taking the square roots of all the parts, affords:

(x + 1) < √(x + 1)(x + 2) < (x + 2).

Substituting x = y − 1 for the last two (x + 2) terms, finally affords:

(x + 1) < √(x + 1)(y + 1) < (y + 1)

and proves the chain inequality: x < √xy < y for all x and for y = x + 1.

The sequence generated from √xy, is shown in Table I where the values of x ranging from 1 to 9 and those of x+1 = y from 2 to 10 are multiplied. The value for each √xy, to 5 decimal places, is irrational and each of these values indeed falls between two consecutive natural numbers. In addition, two larger values for x and x+1 are shown in the last row with a d value even closer to 1.

Table I Irrational (Sequence)
x	x+1	√x(x+1)	d
1	2	1.41421
	1.03528
2	3	2.44948
	1.01462
3	4	3.46410
	1.00804
4	5	4.47214
	1.00509
5	6	5.47723
			1.00351
6	7	6.48074
			1.00257
7	8	7.48331
			1.00197
8	9	8.48528
			1.00155
9	10	9.48683
…	…	…	…
99	100	99.49874
			1.0000125
100	101	100.49876

In the next three sections it will be shown that between two consecutive natural numbers there exists an irrational number √(n(x+y). Part IIa and Part IIb will show two proofs by induction and the third will concentrate on the Staircase sequence generated by √(n(x+y).

Irrational Numbers from Adjacent Natural Numbers (Part I)

A Sequence of Irrational Numbers from √xy